The area of the triangle formed by the lines | vedclass

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Question

(A) The given equation is ${y^2} - 9xy + 18{x^2} = 0$. Factoring the quadratic expression: $(y - 6x)(y - 3x) = 0$. So,the two lines are $y = 6x$ and $

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$\frac{27}{4} 	ext{ sq. units}$

Vedclass · Answer

(A) The given equation is ${y^2} - 9xy + 18{x^2} = 0$. Factoring the quadratic expression: $(y - 6x)(y - 3x) = 0$. So,the two lines are $y = 6x$ and $y = 3x$. The third line is $y = 9$. To find the vertices of the triangle,we find the intersection points: $1$. Intersection of $y = 6x$ and $y = 3x$ is $(0, 0)$. $2$. Intersection of $y = 6x$ and $y = 9$ is $x = \frac{9}{6} = \frac{3}{2}$,so point is $(\frac{3}{2}, 9)$. $3$. Intersection of $y = 3x$ and $y = 9$ is $x = \frac{9}{3} = 3$,so point is $(3, 9)$. The area of the triangle with vertices $(0, 0)$,$(\frac{3}{2}, 9)$,and $(3, 9)$ is given by: $	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ $	ext{Area} = \frac{1}{2} |0(9 - 9) + \frac{3}{2}(9 - 0) + 3(0 - 9)|$ $	ext{Area} = \frac{1}{2} |0 + \frac{27}{2} - 27| = \frac{1}{2} |-\frac{27}{2}| = \frac{27}{4} 	ext{ sq. units}$.

The area of the triangle formed by the lines ${y^2} - 9xy + 18{x^2} = 0$ and $y = 9$ is

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